Intersection of Powers of a Contraction Map is a Single Point

Question

In Elementary Clasical Analysis, (Marsden) (page 318, #14)

Let $f:X \rightarrow X$ be a contraction on a compact metric space $X$. Show that $\bigcap_{n=1}^\infty f^n(X)$ is a single point, where $f^n = f \circ f \circ \cdot \cdot \cdot \circ f$ ($n$ times). Is this true if $X = \mathbb{R}$?

I can prove the first part, using this discussion to show that $\exists \; x_0 \in \bigcap_{n=1}^\infty f^n(X)$. (The uniqueness proof is similarly straightforward).

The second part:

Is this true if $X = \mathbb{R}$?

This seems completely obvious because $\mathbb{R}$ is a compact metric space. But I don't understand why he would have bothered to ask if it was really that simple? Am I missing something?

What makes you think that $\mathbb R$ is compact? – Kavi Rama Murthy Jan 30 '20 at 05:44 — Kavi Rama Murthy, Jan 30 '20 at 05:44
oops, I was missing something. thanks – Peter_Pan Jan 30 '20 at 05:46 — Peter_Pan, Jan 30 '20 at 05:46

score 2 · Accepted Answer · answered Jan 30 '20 at 05:49

2

Let $X=\mathbb R$ and $f(x)=\frac x 2$. Then $f$ is a contraction mapping and $f^{n} (X)=X$ for all $n$ so $\cap_n f^{n} (X) =\mathbb R$!.

answered Jan 30 '20 at 05:49

Kavi Rama Murthy

359,332

Intersection of Powers of a Contraction Map is a Single Point

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